How many multiples of 3 are between 10 and 100? (SAT math question)

Question

In the figure above, circular region A represents all integers from 10 to 100, inclusive; circular region B represents all integers that are multiples of 3; and circular region C represents all squares of integers. How many numbers are represented by the shaded region?

• a) 24
• b) 25
• c) 26
• d) 27
• e) 28

Here's my train of thought: All we need to know is this set of numbers is from 10-100 (which I'm assuming does include 10 and 100?) and is multiples of 3. Instead of listing out every multiple of 3, I can have the largest multiple of 3 less than or equal to 100, which is 99, subtract the smallest multiple of 3 more than or equal to 10, which is 12. 99 - 12 = 87 Then to find the number of times 3 fits into 87, divide 87 by 3. 87 / 3 = 29 Agh! 29 is not one of the answer options provided!

Assuming what I've done so far is correct, now I'm thinking that for some reason 1 has to be subtracted from 29 29 - 1 = 28 to get 28, which is one of the options provided and also is the correct answer.

But why would I need to subtract 1 from 29? (Or is that not the right way to find the answer?)

Answer 1

There are $30$ multiples of three between $10$ and $100$ (and it doesn't matter whether we include $10$ and $100$). You are correct that they range from $12$ through $99$, but you made a fencepost error by not adding one. So there are $30$ numbers in the shaded part plus the center of the diagram. The center has numbers that are between $10$ and $100$, squares, and multiples of $3$ (and hence $9$ as they are squares). These are $36$ and $81$. Subtracting those two gets you to $28$.

Answer 2

$$\left\lfloor\frac{100}3\right\rfloor-\left\lfloor\frac{10}3\right\rfloor$$

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Alternatively, the first term is $12,$ the last being $99$ forming an Arithmetic Series with Common Difference $=3$

So, the $n$th term $=99=12+(n-1)3$

Answer 3

There are $33$ (positive) multiples of $3$ less than $100$, but we need to throw out $3$, $6$, and $9$, so that takes us down to $30$. However, the problem also wants us to throw out any squares that are multiples of $3$, which means square multiples of $9$, which means $9$, $36$, and $81$. We've already thrown out $9$, so we need only throw out the other two, leaving the count at $28$.